Timeline for When is the following fraction an integer $\frac{3^a}{2^b-3}[(\frac{2^b}{3})^c - 1]$ where $a,b,c \in \mathbb{Z}$? [closed]

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Jan 25, 2018 at 11:32	history	closed	YCor abx András Bátkai Vladimir Dotsenko David Handelman		Not suitable for this site
Jan 25, 2018 at 10:52	comment	added	Vladimir Dotsenko		It's equal to $3^{a-c}\frac{(2^b)^c-3^c}{2^b-3}$. Note that $\frac{(2^b)^c-3^c}{2^b-3}$ is always an integer (moreover, $\frac{a^k-b^k}{a-b}$ is always an integer for integer $a,b$), and the numerator is coprime to $3$, so this expression is an integer iff $3^{a-c}$ is an integer. Please use math.stackexchange.com for questions of this level.
Jan 25, 2018 at 9:55	answer	added	jarauh		timeline score: 3
Jan 25, 2018 at 8:31	comment	added	M. Dus		Don't you have any conditions on $a,b,c$ ? I mean, take $a=0$, $b=1$ and $c=1$, you get $\frac{1}{3}$ if I'm not mistaken.
S Jan 25, 2018 at 7:14	history	suggested	Desiderius Severus	CC BY-SA 3.0	Typos in the title
Jan 25, 2018 at 6:32	review	Close votes
Jan 25, 2018 at 11:32
Jan 25, 2018 at 6:25	review	Suggested edits
S Jan 25, 2018 at 7:14
Jan 25, 2018 at 6:16	review	First posts
Jan 25, 2018 at 6:25
Jan 25, 2018 at 6:11	history	asked	mojojojo	CC BY-SA 3.0